點分治入門

重心分解(Centroid Decomposition)

首先要找到樹的中心，樹的重心的定義是：刪除該中心結點得到的最大子樹的頂點數最少的頂點就是樹的重心。運用dfs的方法很容易實現，代碼爲：

void get_hvy(int u, int fa){
    siz[u] = 1, maxx[u] = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(v != fa && !vis[v]){
            get_hvy(v, u);
            siz[u] += siz[v];
            maxx[u] = max(maxx[u], siz[v]);
        }
    }
    maxx[u] = max(maxx[u], S-siz[u]);
    if(!hvy || maxx[hvy] > maxx[u]) hvy = u;
}

對於$poj1741$題
根據重心$s$把樹分解爲若干子樹，那麼所要求的頂點對(計算樹上頂點對的距離)則分爲以下三類：

1. 頂點$u,v$屬於同一子樹的頂點對$(u,v)$
2. 頂點$u,v$不屬於同一子樹的頂點對$(u,v)$
3. 頂點$s$和其他頂點$v$組成的頂點對$(s,u)$
   那麼通過當前重心$s$求的到其他頂點的距離中包括第一種情況，但是這樣會造成重複統計，因此我們需要在對應子樹下面重新計算一邊並進行去重操作。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4+100;

struct Edge{
    int to, val, next;
}e[maxn<<2];

int head[maxn], tot;
int siz[maxn], maxx[maxn], vis[maxn], hvy, S;
int dep[maxn], now[maxn], cnt, ans;
int n, k;

void addedge(int from, int to, int val){
    e[tot].to = to;
    e[tot].val = val;
    e[tot].next = head[from];
    head[from] = tot++;
}

void get_hvy(int u, int fa){
    siz[u] = 1;
    maxx[u] = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(v != fa && !vis[v]){
            get_hvy(v, u);
            siz[u] += siz[v];
            maxx[u] = max(maxx[u], siz[v]);
        }
    }
    maxx[u] = max(maxx[u], S-siz[u]);
    if(!hvy || maxx[u] < maxx[hvy]) hvy = u;
}

void get_dep(int u, int fa){
    now[cnt++] = dep[u];
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(v != fa && !vis[v]){
            dep[v] = dep[u] + e[i].val;
            get_dep(v, u);
        }
    }
}

int get_sum(int u, int dst){
    dep[u] = dst; cnt = 0;
    get_dep(u, -1);
    sort(now, now+cnt);
    int res = 0;
    for(int l = 0, r = cnt-1; l < r; l++){
        while(l < r && now[l] + now[r] > k) r--;
        res += max(0, r-l);
    }
    return res;
}

void get_ans(int u, int fa){
    vis[u] = 1;
    ans += get_sum(u, 0);
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(!vis[v] && v != fa){
            ans -= get_sum(v, e[i].val);
            hvy = 0, S = siz[v];
            get_hvy(v, u);
            get_ans(hvy, -1);
        }
    }
}

int main(){
    while(scanf("%d%d", &n, &k) != EOF){
        if(!n && !k) break;
        memset(head, -1, sizeof(head));
        memset(vis, 0, sizeof(vis));
        tot = 1;
        for(int i = 1; i < n; i++){
            int from, to, val; scanf("%d%d%d", &from, &to, &val);
            addedge(from, to, val);
            addedge(to, from, val);
        }
        hvy = 0; S = n; ans = 0;
        get_hvy(1, -1);
        get_ans(hvy, -1);
        printf("%d\n", ans);
    }
    return 0;
}

對於洛谷OJ P3806 【模板】點分治1
這道題同樣需要對重心分解，但是對於當前重心$s$的所有子樹，$v_1,..,v_n$，當到子樹$v_t$時，$v_1,..,v_{t-1}$到$s$的距離全部用數組judge以下標的形式存儲下來，rem數組則存儲了$v_t$到$s$的距離，這時遍歷m次需要判斷judge[query[i]-rem[j]]是否存在，如果存在，那麼路徑長度爲query[i]的長度存在的。具體看代碼:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4+10;

struct Edge{
    int to, val, next;
}e[maxn<<1];

int query[maxn], ret[maxn];
int head[maxn], tot;
int siz[maxn], maxx[maxn], hvy, S;
int vis[maxn], rem[maxn], dist[maxn], judge[10000007];
int n, m, q[maxn];

void addedge(int from, int to, int val){
    e[tot].to = to;
    e[tot].val = val;
    e[tot].next = head[from];
    head[from] = tot++;
}

// find grativity
void get_hvy(int u, int fa){
    siz[u] = 1, maxx[u] = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(v != fa && !vis[v]){
            get_hvy(v, u);
            siz[u] += siz[v];
            maxx[u] = max(maxx[u], siz[v]);
        }
    }
    maxx[u] = max(maxx[u], S-siz[u]);
    if(!hvy || maxx[hvy] > maxx[u]) hvy = u;
}

void get_dist(int u, int fa){
    rem[++rem[0]] = dist[u];
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(!vis[v] && v != fa){
            dist[v] = dist[u] + e[i].val;
            get_dist(v, u);
        }
    }
}

void get_ans(int u){
    int p = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(vis[v]) continue;
        rem[0] = 0, dist[v] = e[i].val;
        get_dist(v, u); // 處理u的每個子樹

        for(int j = rem[0]; j; --j)
            for(int k = 1; k <= m; k++)
                if(query[k] >= rem[j]) ret[k] |= judge[query[k]-rem[j]];
        for(int j = rem[0]; j; --j)
            q[++p] = rem[j], judge[rem[j]] = 1;
    }
    for(int i = 1; i <= p; i++)
        judge[q[i]] = 0;
}

void cal(int u){
    vis[u] = judge[0] = 1; get_ans(u);
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(vis[v]) continue;
        S = siz[v], hvy = 0;
        get_hvy(v, u); cal(hvy);
    }
}

int main(){
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof(head)); tot = 0;
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i < n; i++){
        int from, to, val; scanf("%d%d%d", &from, &to, &val);
        addedge(from, to, val);
        addedge(to, from, val);
    }
    for(int i = 1; i <= m; i++) scanf("%d", &query[i]);
    S = n; hvy = 0;
    get_hvy(1, -1); cal(hvy);

    for(int i = 1; i <= m; i++)
        if(ret[i]) puts("AYE");
        else puts("NAY");
    return 0;
}